Identities and proofs: practice problems

The binomial theorem and the general term

To find a particular coefficient, write the general term $\binom{n}{r}a^{n-r}b^r$ and choose the $r$ that gives the required power.

Example: in $(2x - 1)^5$, the $x^3$ term comes from $r = 2$, so its coefficient is $\binom{5}{2} \cdot 2^3 \cdot (-1)^2 = 80$.

The same idea handles $\left(x^2 + \frac{1}{x}\right)^6$ (write the general term as a power $x^{12 - 3r}$) and remainders such as $11^{50} \bmod 100$, using $11^{50} = (10 + 1)^{50}$.

Polynomial division and rational expressions

If dividing $A$ by $B$ gives quotient $Q$ and remainder $R$, then

Use this relation to work backwards when the divisor or the quotient is unknown.

Rational expressions work like fractions: factor and cancel, and use a common denominator to add or subtract.

Finding the coefficients of an identity

An equation that holds for every value of $x$ is an identity in $x$. There are two standard methods:

• Compare coefficients of equal powers after expanding both sides.
• Substitute convenient values of $x$ (and check the result at the end).

Example: from $\frac{5x - 1}{(x + 1)(x - 2)} = \frac{a}{x + 1} + \frac{b}{x - 2}$ we get $5x - 1 = a(x - 2) + b(x + 1)$. Putting $x = -1$ gives $a = 2$, and $x = 2$ gives $b = 3$ (partial fractions).

AM-GM and proving inequalities

For $x > 0$, $x + \frac{9}{x} \geq 2\sqrt{9} = 6$, with equality when $x = 3$, so the minimum is $6$.

To prove an inequality, rewrite LHS − RHS as a sum of squares, e.g. $a^2 + b^2 - 2(a + b - 1) = (a - 1)^2 + (b - 1)^2 \geq 0$.